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\(\frac{1}{337.291}+\frac{583}{291}-\frac{2}{337.291}=\frac{583}{291}-\frac{1}{337.291}=\frac{583.337-1}{337.291}=\frac{169652}{98067}\)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
Chúc bạn học tốt !!!
Trả lời:
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\frac{8}{9}\)
\(=0\)
\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
\(\frac{\left(1.3.2.4.3.5......\left(n-2\right)\left(n\right)\left(n-1\right)\left(n+1\right)\right)}{2.2.3.3.4.4...n.n}=\frac{\left(n+1\right)}{2.n}\)
Gọi \(S=\frac{2009}{1}+\frac{2008}{2}+...+\frac{1}{2009}\)
\(\Rightarrow S=\frac{2010-1}{1}+\frac{2010-2}{2}+...+\frac{2010-2009}{2009}\)
\(\Rightarrow S=2010-1+\frac{2010}{2}-1+...+\frac{2010}{2009}-1\)
\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-\left(1+1+..+1\right)\)
\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-2009\)
\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+...+\frac{2010}{2009}+1\)
\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+..+\frac{2010}{2009}+\frac{2010}{2010}\)
\(\Rightarrow S=2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)\)
Khi đó \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}}{2010\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)}=\frac{1}{2010}\)
C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9
C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9
C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9
C = 20/57 + -28/36 + 10/15 + -2/9
C = 20/57 + -7/9 + 2/3 + -2/9
C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )
C = 58/57 + -1
C = 1/57
D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41
D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41
D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41
D = 1 + -1 + 1/41
D = 1/41
E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7
E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7
E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7
E = -1 + 5/7 + 1/257 + 2/7
E = -1 + ( 5/7 + 2/7 ) + 1/127
E = -1 + 1 + 1/127
E = 1/127
Theo đề ta có: \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
\(=\frac{\left(1+2+3...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
= 0